使用 T-SQL 解析 JSON

Posted 2023-05-08

【中文标题】使用 T-SQL 解析 JSON

【英文标题】：Parsing JSON with T-SQL

【发布时间】：2019-08-08 07:48:53

【问题描述】：

我希望只选择 JSON 字符串中的每个值名称和值，并转入 SQL 中的单独列，以便我可以轻松地将它们传递到 powershell 字符串以发送到外部 API

DECLARE @json NVARCHAR(MAX)

SET @json='  "InsertRecordData":   "data":   "AdditionalData":   "DataObjects":   "ObjData":   "Name": "coll_exclude",  "Fields":   "FieldData1":   "Name": "agreement",  "Value": "1234"  ,  "FieldData2":   "Name": "system",  "Value": "live"  ,  "FieldData3":   "Name": "date_added",  "Value": "2019-08-01"  ,  "FieldData4":   "Name": "time_added",  "Value": "11:20"                ';

SELECT *
FROM OPENJSON(@json,'$.InsertRecordData.data.AdditionalData.DataObjects.ObjData.Fields')

所以我想看看

Agreement   System    Date_added   time_added

1234           live      2019-08-01   11:20

【参考方案1】：

使用 mssql pivot()

DECLARE @json NVARCHAR(MAX)

SET @json='  "InsertRecordData":   "data":   "AdditionalData":   "DataObjects":   "ObjData":   "Name": "coll_exclude",  "Fields":   "FieldData1":   "Name": "agreement",  "Value": "1234"  ,  "FieldData2":   "Name": "system",  "Value": "live"  ,  "FieldData3":   "Name": "date_added",  "Value": "2019-08-01"  ,  "FieldData4":   "Name": "time_added",  "Value": "11:20"                ';


select [agreement], [system], [date_added], [time_added] from
(
    SELECT json_value(js.value, '$.Name') as Titles, json_value(js.value, '$.Value') as val
    FROM OPENJSON(@json,'$.InsertRecordData.data.AdditionalData.DataObjects.ObjData.Fields') as js) as SourceTb
PIVOT
(
    max(val)
    FOR Titles in ([agreement], [system], [date_added], [time_added])
) as PivotTable

【讨论】：

太好了.. 谢谢